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Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably component. For each $P\in \mathbb{Q}[t]$, we have the universal family $U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$. Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. Then for any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form Corollary.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably many components. For each $P\in \mathbb{Q}[t]$, we have the universal family $$U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$$ Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. For any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form the Corollary above.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably component. For each $P\in \mathbb{Q}[t]$, we have the universal family $U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$. Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our vairety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. Then for any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form Corollary.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably component. For each $P\in \mathbb{Q}[t]$, we have the universal family $U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$. Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. Then for any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form Corollary.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projecitve varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-dim T$$

proof: For $m\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $dimZ=\kappa(X)$. Resolving the indeterminent locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq dim(\phi(X_{t}))\\ &dimX_{t}-dim(X/Z)\\ &=dim Z-dim T\\ &=\kappa(X)-dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projecitve varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-dim T \\ &=dimX-dimT\\ &=dimX_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $dim Z=dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert schme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably component. For each $P\in \mathbb{Q}[t]$, we have the universal family $U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$. Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our vairety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. Then for any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal faminly $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form Corollary.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably component. For each $P\in \mathbb{Q}[t]$, we have the universal family $U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$. Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our vairety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. Then for any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form Corollary.